/*
  观察式子，我们可以知道f2(f1(x))=k2(k1*x+b1)*b2，
  展开式子得：ans=k2*k1*x+k2*b1*b2，所以新的k=k2*k1，b=k2*b1*b2，
  利用这个原理建立两棵线段树，一棵维护k，一棵维护b。
*/
#include<cstdio>
#include<iostream>
#define N 200010
#define mod 1000000007
#define ll long long
#define lson l,mid,now*2
#define rson mid+1,r,now*2+1
#define lc now*2
#define rc now*2+1
#ifdef unix
#define LL "%lld\n"
#else
#define LL "%I64d\n"
#endif
using namespace std;
ll K[N*4],B[N*4];int n,m;
int read()
{
    int num=0,flag=1;char c=getchar();
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)flag=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){num=num*10+c-‘0‘;c=getchar();}
    return num*flag;
}
void insert(int l,int r,int now,int pos,int k,int b)
{
    if(l==r)
    {
        K[now]=(ll)k;B[now]=(ll)b;return;
    }
    int mid=(l+r)/2;
    if(pos<=mid)insert(lson,pos,k,b);
    else insert(rson,pos,k,b);
    K[now]=K[lc]*K[rc]%mod;B[now]=(K[rc]*B[lc]%mod+B[rc])%mod;
}
ll kquery(int l,int r,int now,int x,int y)
{
    if(l>=x&&r<=y)return K[now];
    int mid=(l+r)/2;
    ll k1=1,k2=1;
    if(x<=mid)k1=kquery(lson,x,y);
    if(y>mid)k2=kquery(rson,x,y);
    return k1*k2%mod;
}
ll bquery(int l,int r,int now,int x,int y)
{
    if(l>=x&&r<=y)return B[now];
    int mid=(l+r)/2;
    ll b1=0,b2=0;
    if(x<=mid)b1=bquery(lson,x,y);
    if(y>mid)
    {
        b2=bquery(rson,x,y);
        b1*=kquery(rson,x,y);
    }
    return (b1%mod+b2)%mod;
}
int main()
{
    freopen("fx.in","r",stdin);
    freopen("fx.out","w",stdout);
    n=read(),m=read();
    for(int i=1;i<=n;i++)
    {
        int k=read(),b=read();
        insert(1,n,1,i,k,b);
    }
    for(int i=1;i<=m;i++)
    {
        char c;cin>>c;
        if(c==‘M‘)
        {
            int p=read(),k=read(),b=read();
            insert(1,n,1,p,k,b);
        }
        else
        {
            int l=read(),r=read(),x=read();
            ll k=kquery(1,n,1,l,r);
            ll b=bquery(1,n,1,l,r);
            printf(LL,(k*x%mod+b)%mod);
        }
    }
    return 0;
}