#include <bits/stdc++.h>

using namespace std;
const int MAXN=3005,MOD=1e9+7;
char s[MAXN];
int n,m,x,y,pre[MAXN],lst[MAXN],dp[MAXN][MAXN];

int main()
{
	scanf("%d%d%s",&n,&m,s+1);s[n+1]='0';
	for(int i=1;i<=n+1;i++) 
		pre[i]=(s[i]-'0')+pre[i-1],lst[i]=i;
	for(int i=1;i<=m;i++)
		scanf("%d%d",&x,&y),lst[x]=max(lst[x],y);
	for(int i=1;i<=n+1;i++) lst[i]=max(lst[i],lst[i-1]);
	
	dp[1][pre[lst[1]]]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=n;j++)
			if(dp[i][j])
			{
				int l=lst[i]+1,r=lst[i+1];
				if(j) (dp[i+1][pre[r]-pre[l-1]+j-1]+=dp[i][j])%=MOD;
				if(lst[i]-(i-1)-j) (dp[i+1][pre[r]-pre[l-1]+j]+=dp[i][j])%=MOD;
			}
	printf("%d",dp[n+1][0]);
	return 0;
}
### Prompt

In CPP, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;
const int MAXN=3005,MOD=1e9+7;
char s[MAXN];
int n,m,x,y,pre[MAXN],lst[MAXN],dp[MAXN][MAXN];

int main()
{
	scanf("%d%d%s",&n,&m,s+1);s[n+1]='0';
	for(int i=1;i<=n+1;i++) 
		pre[i]=(s[i]-'0')+pre[i-1],lst[i]=i;
	for(int i=1;i<=m;i++)
		scanf("%d%d",&x,&y),lst[x]=max(lst[x],y);
	for(int i=1;i<=n+1;i++) lst[i]=max(lst[i],lst[i-1]);
	
	dp[1][pre[lst[1]]]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=n;j++)
			if(dp[i][j])
			{
				int l=lst[i]+1,r=lst[i+1];
				if(j) (dp[i+1][pre[r]-pre[l-1]+j-1]+=dp[i][j])%=MOD;
				if(lst[i]-(i-1)-j) (dp[i+1][pre[r]-pre[l-1]+j]+=dp[i][j])%=MOD;
			}
	printf("%d",dp[n+1][0]);
	return 0;
}
```