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 int P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 10;
int n, m, cnt;
vector<int> g[N];
int dfn[N], fa[N], L[N], R[N];
void dfs(int first) {
  dfn[first] = ++*dfn;
  for (int second : g[first]) {
    if (dfn[second]) {
      if (dfn[second] < dfn[first] || second == first) continue;
      fa[first] = second, ++cnt;
      L[cnt] = R[cnt] = first;
      for (; second != first; second = fa[second]) {
        L[cnt] = min(L[cnt], second);
        R[cnt] = max(R[cnt], second);
      }
    } else
      fa[second] = first, dfs(second);
  }
}
int f[N];
long long sum[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v), g[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i)
    if (!dfn[i]) dfs(i);
  for (int i = 1; i <= cnt; ++i) f[R[i]] = max(f[R[i]], L[i]);
  for (int i = 1; i <= n; ++i) {
    f[i] = max(f[i], f[i - 1]);
    sum[i] = sum[i - 1] + (i - f[i]);
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    int pos = lower_bound(f + 1, f + 1 + n, l) - f;
    if (pos > r) {
      printf("%lld\n", (long long)(r - l + 1) * (r - l + 2) / 2);
    } else {
      long long ans = sum[r] - sum[pos - 1];
      long long d = pos - l;
      printf("%lld\n", ans + d * (1 + d) / 2);
    }
  }
}
