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 int N = 3e3 + 5, MOD = 1e9 + 7;
int dp[N][N], a[N];

int mul(int a, int b) {return 1ll * a * b % MOD;}
int Pow(int a, int b) {
    int res = 1;
    for (; b; b>>=1, a = mul(a, a)) if (b&1) res = mul(res, a);
    return res;
}
int Inv = Pow(2, MOD - 2);

int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, q; cin >> n >> q;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) dp[i][j] = a[i] > a[j];
    int ans = 1;
    for (int j = 0; j < q; j++) {
        int x, y; cin >> x >> y;
        for (int i = 1; i <= n; i++) if (i != x && i != y) {
			dp[i][x] = dp[i][y] = mul(dp[i][x] + dp[i][y], Inv);
			dp[x][i] = dp[y][i] = mul(dp[x][i] + dp[y][i], Inv);
		}
        dp[x][y] = dp[y][x] = mul(dp[x][y] + dp[y][x], Inv);
        ans = mul(2, ans);
    }
    int sum = 0;
    for (int i = 1; i <= n; i++) for (int j = i + 1; j <= n; j++) sum = (sum + dp[i][j]) % MOD;
    cout << mul(ans, sum) << "\n";


    return 0;
}