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;

const int64_t MOD = 1e9+7;
void add(int64_t& a, int64_t b){
    a = (a+b) % MOD;
}

int64_t power_mod(int64_t num, int64_t power){
    int64_t prod = 1;
    num %= MOD;
    while(power > 0){
        if(power&1) prod = prod * num % MOD;
        num = num * num % MOD;
        power >>= 1;
    }
    return prod;
}

int64_t dp[3000][3000];

int main(){
    int N, Q, A[3000];
    cin >> N >> Q;
    for(int i=0; i<N; i++) cin >> A[i];

    for(int i=0; i<N; i++) for(int j=0; j<N; j++) if(A[i] > A[j]) dp[i][j] = 1;
    int64_t inv2 = power_mod(2, MOD-2);

    for(int q=0; q<Q; q++){
        int x, y;
        cin >> x >> y;
        x--; y--;

        for(int i=0; i<N; i++) if(i!=x && i!=y){
            dp[x][i] = dp[y][i] = (dp[x][i] + dp[y][i]) * inv2 % MOD;
            dp[i][x] = dp[i][y] = (dp[i][x] + dp[i][y]) * inv2 % MOD;
        }
        dp[x][y] = dp[y][x] = (dp[x][y] + dp[y][x]) * inv2 % MOD;
    }

    int64_t ans = 0;
    for(int i=0; i<N; i++) for(int j=i+1; j<N; j++) add(ans, dp[i][j]);
    ans = ans * power_mod(2, Q) % MOD;
    cout << ans << endl;
    return 0;
}