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;

#define SZ(x) (int)(x).size()
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define RFOR(i, a, b) for (int i=a;i>=b;--i)

const int MX_N = 3005;
const int MX_Q = 3005;
const int MOD = 1e9+7;

int N, Q, A[MX_N];

int dp[MX_N][MX_N];

int expo(int b, int p) {
    int x = 1;
    while (p > 0) {
        if (p&1) x = 1LL * x * b % MOD;
        b = 1LL * b * b % MOD;
        p >>= 1;
    }
    return x;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> N >> Q;
    FOR(i,1,N){
        cin >> A[i];
    }

    FOR(i,1,N) FOR(j,1,N) dp[i][j] = A[i] > A[j];

    long long half = expo(2,MOD-2), all = expo(2,Q);
    FOR(i,1,Q){
        int X, Y; cin >> X >> Y;

        dp[X][Y] = dp[Y][X] = (half * dp[X][Y] % MOD + half * dp[Y][X] % MOD) % MOD;
        FOR(Z,1,N) if (Z != X and Z != Y) {
            dp[X][Z] = dp[Y][Z] = (half * dp[X][Z] % MOD + half * dp[Y][Z] % MOD) % MOD;
            dp[Z][X] = dp[Z][Y] = (half * dp[Z][X] % MOD + half * dp[Z][Y] % MOD) % MOD;
        }
    }

    int ans = 0;
    FOR(i,1,N) FOR(j,i+1,N) ans = (ans + dp[i][j] * all % MOD) % MOD;
    cout << ans;
}