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 pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()
#define dump(x) cout << #x << " = " << (x) << endl

const ll MOD = 1000000007;
const ll i2 = (MOD + 1) / 2;
const int MN = 3010;

int N, Q;
int A[MN];
int X[MN], Y[MN];
ll p[MN][MN]; // <

int main() {
	cin >> N >> Q;
	rep(i, N) cin >> A[i];
	rep(i, N) {
		rep(j, N) {
			if (A[i] < A[j]) {
				p[i][j] = 1;
			}
		}
	}

	rep(i, Q) {
		cin >> X[i] >> Y[i];
		--X[i]; --Y[i];
	}

	rep(n, Q) {
		int a = X[n], b = Y[n];

		ll t1 = (p[a][b] + p[b][a]) * i2 % MOD;
		p[a][b] = p[b][a] = t1;

		rep(i, N) if (i != a && i != b) {
			ll t = (p[i][a] + p[i][b]) * i2 % MOD;
			p[i][a] = p[i][b] = t;
			t = (p[a][i] + p[b][i]) * i2 % MOD;
			p[a][i] = p[b][i] = t;
		}

	}

	ll ans = 0;
	rep(i, N) {
		rep(j, i) {
			ans = (ans + p[i][j]) % MOD;
		}
	}

	rep(i, Q) ans = ans * 2 % MOD;

	cout << ans << endl;

	return 0;
}