Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long i64;
inline int rd(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 5e3+5, P = 1e9+7, I2 = (P+1)/2;
int n, m, a[N], f[N][N], ans, S; void fix(int &x) { x<P?0:x-=P; }
int main()
{
    n = rd(), m = rd(), S = 1;
    for(int i = 1; i <= n; ++i) a[i] = rd();
    for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) f[i][j] = a[i] > a[j];
    for(int i = 1; i <= m; ++i) 
    {
        int x = rd(), y = rd();
        for(int j = 1; j <= n; ++j) if(x != j&&y != j)
            f[j][x] = f[j][y] = (i64)I2*(f[j][x]+f[j][y])%P,
            f[x][j] = f[y][j] = (i64)I2*(f[x][j]+f[y][j])%P;
        f[x][y] = f[y][x] = (i64)I2*(f[x][y]+f[y][x])%P, fix(S<<=1);
    }
    for(int i = 1; i <= n; ++i) for(int j = i+1; j <= n; ++j) fix(ans += f[i][j]);
    printf("%lld\n", (i64)ans*S%P);
    return 0;
}