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 ll;
const int mod = (int) 1e9 + 7;

ll modPow(ll n, ll k) {
	ll res = 1;
	for (n %= mod; k; k >>= 1) {
		if (k & 1)
			res = res * n % mod;
		n = n * n % mod;
	}
	return res;
}

ll modInv(ll x) {
	return modPow(x, mod - 2);
}

const int MaxN = 3002;

const ll Half = modInv(2);
ll dp[MaxN + 1][MaxN + 1];

void avg(ll &x, ll &y) {
	x = y = (x + y) % mod * Half % mod;
}

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

	int n, Q;
	cin >> n >> Q;

	{
		static ll a[MaxN + 9];
		for (int i = 0; i < n; ++i)
			cin >> a[i];
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n; ++j)
				dp[i][j] = a[i] > a[j];
	}

	for (int rep = 0; rep < Q; ++rep) {
		int i, j;
		cin >> i >> j;
		--i, --j;

		avg(dp[i][j], dp[j][i]);

		for (int k = 0; k < n; ++k)
			if (k != i && k != j) {
				avg(dp[i][k], dp[j][k]);
				avg(dp[k][i], dp[k][j]);
			}
	}

	ll ans = 0;
	for (int i = 0; i < n; ++i)
		for (int j = i + 1; j < n; ++j)
			(ans += dp[i][j]) %= mod;

	ans *= modPow(2, Q);

	cout << (ans % mod + mod) % mod;
	return 0;
}