/*
神题
朴素DP-->O(n^3)
DP：f[i][j]=填完了前i-1个数，当前需要考虑的区间[i,R]中需要填入j个1，方案数
ans=f[n+1][0]
转移枚举i位置的数即可O(1)转移
O(n^2)
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 3010
#define mod 1000000007
using namespace std;
int n, m, x, y, f[N][N], s[N], sum, R[N];
char S[N];
int main(){
	scanf("%d%d%s", &n, &m, S+1);
	s[0]=0; for(int i=1; i<=n; i++)s[i]=s[i-1]+S[i]-'0'; s[n+1]=s[n];
	for(int i=1; i<=n+1; i++)R[i]=i;
	for(int i=1; i<=m; i++){scanf("%d%d", &x, &y); R[x]=max(R[x], y);}
	memset(f, 0, sizeof(f)); f[1][s[R[1]]]=1; x=R[1];
	for(int i=1; i<=n; i++){
		y=max(x, R[i+1]); sum=s[y]-s[x];
		for(int j=0; j<=x-i+1; j++){
			if(j)f[i+1][j-1+sum]=(f[i+1][j-1+sum]+f[i][j])%mod;
			if(j<=x-i)f[i+1][j+sum]=(f[i+1][j+sum]+f[i][j])%mod;
		}
		x=y;
	}
	printf("%d", f[n+1][0]);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
/*
神题
朴素DP-->O(n^3)
DP：f[i][j]=填完了前i-1个数，当前需要考虑的区间[i,R]中需要填入j个1，方案数
ans=f[n+1][0]
转移枚举i位置的数即可O(1)转移
O(n^2)
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 3010
#define mod 1000000007
using namespace std;
int n, m, x, y, f[N][N], s[N], sum, R[N];
char S[N];
int main(){
	scanf("%d%d%s", &n, &m, S+1);
	s[0]=0; for(int i=1; i<=n; i++)s[i]=s[i-1]+S[i]-'0'; s[n+1]=s[n];
	for(int i=1; i<=n+1; i++)R[i]=i;
	for(int i=1; i<=m; i++){scanf("%d%d", &x, &y); R[x]=max(R[x], y);}
	memset(f, 0, sizeof(f)); f[1][s[R[1]]]=1; x=R[1];
	for(int i=1; i<=n; i++){
		y=max(x, R[i+1]); sum=s[y]-s[x];
		for(int j=0; j<=x-i+1; j++){
			if(j)f[i+1][j-1+sum]=(f[i+1][j-1+sum]+f[i][j])%mod;
			if(j<=x-i)f[i+1][j+sum]=(f[i+1][j+sum]+f[i][j])%mod;
		}
		x=y;
	}
	printf("%d", f[n+1][0]);
	return 0;
}
```