You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 10, MAX = 3e5 + 1e4, MOD = 1e9 + 7, MAXL = 25;
vector<int> adj[MAX];
long long dp[MAXL][MAX], ma[MAXL][MAX], mi[MAXL][MAX], a[MAX], h[MAX], pa[MAX];
bool vis[MAX];
pair<long long, long long> g(int v, int x) {
  long long mii = INF, maa = 0;
  for (int i = 0; i <= 20; i++) {
    if (x & (1 << i)) {
      mii = min(mii, mi[i][v]);
      maa = max(maa, ma[i][v]);
      v = dp[i][v];
    }
  }
  return {mii, maa};
}
void dfs(int v, int p) {
  vis[v] = 1;
  h[v] = h[p] + 1;
  dp[0][v] = p;
  mi[0][v] = min(v, p);
  ma[0][v] = max(v, p);
  for (int i = 1; i <= 20; i++) {
    dp[i][v] = dp[i - 1][dp[i - 1][v]];
    mi[i][v] = min(mi[i - 1][v], mi[i - 1][dp[i - 1][v]]);
    ma[i][v] = max(ma[i - 1][v], ma[i - 1][dp[i - 1][v]]);
  }
  for (int u : adj[v]) {
    if (!vis[u])
      dfs(u, v);
    else if (u != p && h[v] > h[u]) {
      pair<long long, long long> x = g(v, h[v] - h[u]);
      a[x.first] = min(a[x.first], x.second - 1);
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, q;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= n + 1; i++) a[i] = INF;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, i);
  for (int i = n; i > 0; i--) a[i] = min(a[i], a[i + 1]);
  for (int i = 1; i <= n; i++) pa[i] = pa[i - 1] + a[i] - i + 1;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    long long x = lower_bound(a + l, a + r + 2, r + 1) - a;
    cout << pa[x - 1] - pa[l - 1] + (((r + 1 - x) * (r + 2 - x)) / 2) << "\n";
  }
  return 0;
}
