1 /*
  2 http://acm.hdu.edu.cn/showproblem.php?pid=4893
  3 题意:三个操作 某点增加n, 某段 全部变成 最近最小的斐波那契,查询某段和 
  4 思路:就是个线段树,爆炸
  5 线段树时从下往上更新,我是zz
  6 2017年02月27日20:49:32
  7 */
  8 #include <cstdio>
  9 #include <cstring>
 10 #define N 100010
 11 long long xds[N<<2];
 12 bool xdsisfb[N<<2];
 13 long long fb[92]={
 14 1,
 15 1,
 16 2,
 17 3,
 18 5,
 19 8,
 20 13,
 21 21,
 22 34,
 23 55,
 24 89,
 25 144,
 26 233,
 27 377,
 28 610,
 29 987,
 30 1597,
 31 2584,
 32 4181,
 33 6765,
 34 10946,
 35 17711,
 36 28657,
 37 46368,
 38 75025,
 39 121393,
 40 196418,
 41 317811,
 42 514229,
 43 832040,
 44 1346269,
 45 2178309,
 46 3524578,
 47 5702887,
 48 9227465,
 49 14930352,
 50 24157817,
 51 39088169,
 52 63245986,
 53 102334155,
 54 165580141,
 55 267914296,
 56 433494437,
 57 701408733,
 58 1134903170,
 59 1836311903,
 60 2971215073,
 61 4807526976,
 62 7778742049,
 63 12586269025,
 64 20365011074,
 65 32951280099,
 66 53316291173,
 67 86267571272,
 68 139583862445,
 69 225851433717,
 70 365435296162,
 71 591286729879,
 72 956722026041,
 73 1548008755920,
 74 2504730781961,
 75 4052739537881,
 76 6557470319842,
 77 10610209857723,
 78 17167680177565,
 79 27777890035288,
 80 44945570212853,
 81 72723460248141,
 82 117669030460994,
 83 190392490709135,
 84 308061521170129,
 85 498454011879264,
 86 806515533049393,
 87 1304969544928657,
 88 2111485077978050,
 89 3416454622906707,
 90 5527939700884757,
 91 8944394323791464,
 92 14472334024676221,
 93 23416728348467685,
 94 37889062373143906,
 95 61305790721611591,
 96 99194853094755497,
 97 160500643816367088,
 98 259695496911122585,
 99 420196140727489673,
100 679891637638612258,
101 1100087778366101931,
102 1779979416004714189,
103 2880067194370816120,
104 4660046610375530309,
105 7540113804746346429};
106 long long findfb(long long val){
107     int l=0,r=91,m;
108     while(l!=r){
109         m=(l+r)/2;
110         if(fb[m]<val){
111             l=m+1;
112         }else{
113             r=m;
114         }
115     }
116     if(l!=0 && val-fb[l-1]<=fb[l]-val){
117         return fb[l-1];
118     }else{
119         return fb[l];
120     }
121 }
122 void add(int l,int r,int now,int id,long long value){
123     if(l==r){
124         xdsisfb[now]=false;
125         xds[now]+=value;
126         return;
127     }
128     int m=(l+r)/2;
129     if(m<id){
130         add(m+1,r,now<<1|1,id,value);
131     }else{
132         add(l,m,now<<1,id,value);
133     }
134     xds[now]=xds[now<<1]+xds[now<<1|1];
135     xdsisfb[now]=xdsisfb[now<<1]&&xdsisfb[now<<1|1];
136 }
137 long long query(int l,int r,int now,int zz,int yy){
138     if(zz<=l && yy>=r)return xds[now];
139     int m=(l+r)/2;
140     if(m<zz){
141         return query(m+1,r,now<<1|1,zz,yy);
142     }else if(m>=yy){
143         return query(l,m,now<<1,zz,yy);
144     }else{
145         return query(m+1,r,now<<1|1,zz,yy)+query(l,m,now<<1,zz,yy);
146     }
147 }
148 void gxfb(int l,int r,int now,int zz,int yy){
149     if(xdsisfb[now])return;
150     if(l==r){
151         xdsisfb[now]=true;
152         xds[now]=findfb(xds[now]);
153         return ;
154     }
155     int m=(l+r)/2;
156     if(m<zz){
157         gxfb(m+1,r,now<<1|1,zz,yy);
158     }else if(m>=yy){
159         gxfb(l,m,now<<1,zz,yy);
160     }else{
161         gxfb(m+1,r,now<<1|1,zz,yy);
162         gxfb(l,m,now<<1,zz,yy);
163     }
164     xds[now]=xds[now<<1]+xds[now<<1|1];
165     xdsisfb[now]=xdsisfb[now<<1]&&xdsisfb[now<<1|1];
166 }
167 int main(){
168     //freopen("test.in","r",stdin);
169     //freopen("outb","w",stdout);
170     int n,m;
171     while(scanf("%d %d",&n,&m)!=EOF){
172         memset(xdsisfb,false,sizeof(xdsisfb));
173         memset(xds,0,sizeof(xds));
174         long long a,b,c;
175         while(m--){
176             scanf("%lld %lld %lld",&a,&b,&c);
177             if(a==1){
178                 add(1,n,1,b,c);
179             }else if(a==2){
180                 printf("%lld\n",query(1,n,1,b,c));
181             }else if(a==3){
182                 gxfb(1,n,1,b,c);
183             }
184         }
185     }
186     return 0;
187 }