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 mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
template <class T>
T gcd(T a, T b) {
  if (!b) return a;
  return gcd(b, a % b);
}
const int maxn = 3e5 + 10;
vector<vector<int> > G(maxn);
int dfn[maxn], low[maxn];
int xl[maxn], xr[maxn];
int cnt, dep;
stack<int> sk;
int to[maxn];
long long sum[maxn];
void dfs(int u, int f) {
  dfn[u] = low[u] = ++dep;
  sk.push(u);
  for (int i = 0; i < (int)G[u].size(); i++) {
    int v = G[u][i];
    if (v == f) continue;
    if (!dfn[v]) {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
    } else
      low[u] = min(low[u], dfn[v]);
  }
  if (dfn[u] == low[u]) {
    xl[cnt] = xr[cnt] = u;
    while (sk.top() != u) {
      xl[cnt] = min(xl[cnt], sk.top());
      xr[cnt] = max(xr[cnt], sk.top());
      sk.pop();
    }
    sk.pop();
    cnt++;
  }
}
int main() {
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    if (!dfn[i]) dfs(i, 0);
  }
  for (int i = 1; i <= n; i++) to[i] = n;
  for (int i = 0; i < cnt; i++) {
    if (xl[i] != xr[i]) {
      to[xl[i]] = xr[i] - 1;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    to[i] = min(to[i], to[i + 1]);
  }
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + (to[i] - i + 1);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int l, r, p, L, R;
    scanf("%d", &L);
    scanf("%d", &R);
    l = L, r = R, p = R + 1;
    while (l <= r) {
      int m = (l + r) / 2;
      if (to[m] >= R) {
        p = m;
        r = m - 1;
      } else
        l = m + 1;
    }
    long long ans =
        sum[p - 1] - sum[L - 1] + (long long)(R - p + 2) * (R - p + 1) / 2;
    printf("%lld\n", ans);
  }
  return 0;
}
