1 #include<bits/stdc++.h>
 2 #define F(i,a,b) for(int i=a;i<=b;++i)
 3 using namespace std;
 4 
 5 const int N=2e6+7;
 6 
 7 struct node
 8 {
 9     int op,x,a,b;
10     node(int _a=0,int _b=0,int _c=0,int _d=0):op(_a),x(_b),a(_c),b(_d){}
11     bool operator <(const node &B)const{return x!=B.x?x<B.x:op<B.op;}
12 }A[N];
13 
14 int t,n,m,x,y,a[2000],ed,cnt[2],tp[2];
15 int hsh[N*2],h_ed,C[N];
16 char ans[N];
17 
18 inline void add(int x,int c){while(x<=h_ed)C[x]+=c,x+=x&-x;}
19 inline int ask(int x){int an=0;while(x)an+=C[x],x-=x&-x;return an;}
20 int idx(int x){return lower_bound(hsh+1,hsh+1+h_ed,x)-hsh;}
21 
22 
23 int main(){
24     scanf("%d",&t);
25     while(t--)
26     {
27         ed=0,scanf("%d%d",&n,&m);
28         F(i,1,n)scanf("%d",a+i);
29         F(i,1,n)
30         {
31             cnt[0]=cnt[1]=0;
32             cnt[!(i&1)]+=a[i];
33             tp[0]=tp[1]=0;
34             tp[!(i&1)]+=a[i];
35             A[++ed]=node(-1,cnt[0]-tp[0],cnt[1]-tp[1],cnt[1]);
36             A[++ed]=node(1,cnt[0],cnt[1]-tp[1],cnt[1]);
37             F(j,i+1,n)
38             {
39                 cnt[!(j&1)]+=a[j];
40                 tp[0]=tp[1]=0;
41                 tp[!(i&1)]+=a[i];
42                 tp[!(j&1)]+=a[j];
43                 A[++ed]=node(-1,cnt[0]-tp[0],cnt[1]-tp[1],cnt[1]);
44                 A[++ed]=node(1,cnt[0],cnt[1]-tp[1],cnt[1]);
45             }
46         }
47         F(i,1,m)
48         {
49             scanf("%d%d",&x,&y);
50             A[++ed]=node(0,x,y,i);
51         }
52         sort(A+1,A+1+ed),h_ed=0;
53         F(i,1,ed)
54         {
55             if(A[i].op!=0)hsh[++h_ed]=A[i].a,hsh[++h_ed]=A[i].b;
56             else hsh[++h_ed]=A[i].a;
57         }
58         sort(hsh+1,hsh+1+h_ed),h_ed=unique(hsh+1,hsh+1+h_ed)-hsh-1;
59         F(i,1,ed)
60         {
61             if(A[i].op!=0)A[i].a=idx(A[i].a),A[i].b=idx(A[i].b);
62             else A[i].a=idx(A[i].a);
63         }
64         F(i,1,ed)
65         {
66             if(A[i].op)
67             {
68                 add(A[i].a,-1*A[i].op);
69                 add(A[i].b+1,A[i].op);
70             }else
71             {
72                 int now=ask(A[i].a);
73                 ans[A[i].b]=(now!=0)+‘0‘;
74             }
75         }
76         ans[m+1]=0;
77         printf("%s\n",ans+1);
78         F(i,0,h_ed)C[i]=0;
79     }
80     return 0;
81 }