Problem: There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

Solution (in CPP):

Solution:
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
void read(int &x) {
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define rg register
const int maxn=3010,mod=1e9+7;
char ch[maxn];
int n,m,f[maxn][maxn],tot,sum[maxn],las[maxn];
int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
int main()
{
    read(n),read(m);scanf("%s",ch+1);
    for(rg int i=1;i<=n+1;i++){
        sum[i]=sum[i-1],las[i]=i;
        if(ch[i]=='1')tot++,sum[i]++;
    }
    for(rg int i=1,x,y;i<=m;i++)read(x),read(y),las[x]=max(las[x],y);
    for(rg int i=1;i<=n+1;i++)las[i]=max(las[i],las[i-1]);
    f[1][sum[las[1]]]=1;
    for(rg int i=1;i<=n;i++)
        for(rg int j=0;j<=tot;j++)
            if(f[i][j]){
                int a=las[i],b=las[i+1];
                if(j)f[i+1][j+sum[b]-sum[a]-1]=add(f[i+1][j+sum[b]-sum[a]-1],f[i][j]);
                if(las[i]+1-i-j)f[i+1][j+sum[b]-sum[a]]=(f[i+1][j+sum[b]-sum[a]],f[i][j]);
        }
    printf("%d\n",f[n+1][0]);
}