#include <cstdio>

inline int Min(const int a,const int b){return a<b?a:b;}
inline int Max(const int a,const int b){return a>b?a:b;}

int n,m,Sum[3005],r[3005],f[3005][3005];
char s[3005];
const int p=1000000007;

int main()
{
    scanf("%d%d%s",&n,&m,s+1);
    for(int i=1;i<=n+1;++i)Sum[i]=Sum[i-1]+(s[i]&1),r[i]=i;
    for(int i=1,L,R;i<=m;++i)scanf("%d%d",&L,&R),r[L]=Max(r[L],R);
    for(int i=1;i<=n;++i)r[i]=Max(r[i],r[i-1]);
    for(int i=f[1][Sum[r[1]]]=1;i<=n;++i)
        for(int j=0;j<=n;++j)
            if(f[i][j])
            {
                int L=r[i]+1,R=r[i+1];
                int c1=j,c0=r[i]-i+1-j,Nc=Sum[R]-Sum[L-1];
                if(c0)(f[i+1][j+Nc]+=f[i][j])%=p;
                if(c1)(f[i+1][j+Nc-1]+=f[i][j])%=p;
            }
    printf("%d\n",f[n+1][0]);
    return 0;
}
### Prompt

Create a solution in CPP for 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 <cstdio>

inline int Min(const int a,const int b){return a<b?a:b;}
inline int Max(const int a,const int b){return a>b?a:b;}

int n,m,Sum[3005],r[3005],f[3005][3005];
char s[3005];
const int p=1000000007;

int main()
{
    scanf("%d%d%s",&n,&m,s+1);
    for(int i=1;i<=n+1;++i)Sum[i]=Sum[i-1]+(s[i]&1),r[i]=i;
    for(int i=1,L,R;i<=m;++i)scanf("%d%d",&L,&R),r[L]=Max(r[L],R);
    for(int i=1;i<=n;++i)r[i]=Max(r[i],r[i-1]);
    for(int i=f[1][Sum[r[1]]]=1;i<=n;++i)
        for(int j=0;j<=n;++j)
            if(f[i][j])
            {
                int L=r[i]+1,R=r[i+1];
                int c1=j,c0=r[i]-i+1-j,Nc=Sum[R]-Sum[L-1];
                if(c0)(f[i+1][j+Nc]+=f[i][j])%=p;
                if(c1)(f[i+1][j+Nc-1]+=f[i][j])%=p;
            }
    printf("%d\n",f[n+1][0]);
    return 0;
}
```