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 std::max;
using std::min;
int n, m, q;
int h[(300000 + 5)], tot;
struct Edge {
  int to, next;
} e[(300000 + 5) * 2];
int dfn[(300000 + 5)], low[(300000 + 5)], cnt;
int st[(300000 + 5)], tp;
int nxt[(300000 + 5)];
long long sum[(300000 + 5)];
void Read(int &x) {
  x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
}
void Write(long long x) {
  if (x >= 10) Write(x / 10);
  putchar(x % 10 + '0');
}
void AddEdge(int u, int v) {
  e[++tot].to = v;
  e[tot].next = h[u];
  h[u] = tot;
}
void Tarjan(int u) {
  dfn[u] = low[u] = ++cnt;
  st[++tp] = u;
  int v;
  for (int i = h[u]; i; i = e[i].next) {
    v = e[i].to;
    if (!dfn[v]) {
      Tarjan(v);
      low[u] = min(low[u], low[v]);
      if (dfn[u] <= low[v]) {
        int mn = u, mx = u, tmp = 1;
        while (st[tp] != u) {
          ++tmp;
          mn = min(mn, st[tp]);
          mx = max(mx, st[tp]);
          --tp;
        }
        if (tmp >= 3) nxt[mn] = min(nxt[mn], mx);
      }
    } else
      low[u] = min(low[u], dfn[v]);
  }
}
int main() {
  Read(n), Read(m);
  int u, v;
  for (int i = 1; i <= m; ++i) Read(u), Read(v), AddEdge(u, v), AddEdge(v, u);
  for (int i = 1; i <= n; ++i) nxt[i] = n + 1;
  for (int i = 1; i <= n; ++i)
    if (!dfn[i]) Tarjan(i);
  for (int i = n - 1; i; --i) nxt[i] = min(nxt[i + 1], nxt[i]);
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + nxt[i] - i;
  Read(q);
  int ql, qr, l, r, mid;
  long long ans;
  while (q--) {
    Read(ql), Read(qr);
    l = ql - 1, r = qr;
    while (l < r) {
      mid = l + r + 1 >> 1;
      if (nxt[mid] > qr)
        r = mid - 1;
      else
        l = mid;
    }
    ans = sum[r] - sum[ql - 1] + (long long)(qr - r + 1) * (qr - r) / 2;
    Write(ans), putchar('\n');
  }
  return 0;
}
