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 N = 300100;
struct edge {
  int nxt, to;
} e[N << 1];
struct fe {
  int l, r;
  bool operator<(const fe &o) const { return r < o.r; }
} b[N];
int ft[N], q[N], dfn[N], top, cnt, n, m, num, a[N];
int fa[N];
deque<pair<int, int>> c;
long long s[N];
bool vis[N];
void dfs(int x, int fa) {
  q[top++] = x;
  vis[x] = 1;
  dfn[x] = num++;
  for (int i = ft[x]; i; i = e[i].nxt)
    if (e[i].to != fa) {
      int y = e[i].to;
      if (!vis[y])
        dfs(y, x);
      else if (dfn[y] < dfn[x]) {
        int mn = y, mx = y;
        for (int j = top - 1; q[j] != y; j--) {
          if (mn > q[j]) mn = q[j];
          if (mx < q[j]) mx = q[j];
        }
        b[cnt++] = (fe){mn, mx};
      }
    }
  top--;
}
int getf(int v) { return fa[v] == v ? v : fa[v] = getf(fa[v]); }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[i << 1] = (edge){ft[x], y};
    ft[x] = i << 1;
    e[i << 1 | 1] = (edge){ft[y], x};
    ft[y] = i << 1 | 1;
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 0);
  for (int i = 0; i < cnt; i++) {
    fa[b[i].r] = b[i].l;
  }
  cnt = 0;
  for (int i = 1; i <= n; i++)
    if (fa[i] != 0) b[cnt].l = fa[i], b[cnt].r = i, cnt++;
  int now = 0, last = -1;
  for (int i = 1, j = 0; i <= n; i++) {
    while (j < cnt && i >= b[j].r) now = max(now, b[j].l), j++;
    a[i] = now;
    if (a[i] != 0 && a[i] != last) {
      last = a[i];
      c.push_back(make_pair(a[i], i));
    }
  }
  c.push_back(make_pair(n + 1, n + 1));
  for (int i = 1; i <= n; i++) fa[i] = i;
  int L = c.front().first, R = c.back().second;
  last = 1;
  while (last <= n) {
    if (last <= c.front().first)
      fa[last] = c.front().second + n + 10;
    else
      c.pop_front(), last--;
    last++;
  }
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + i - a[i];
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    long long ans = 0;
    int u;
    if (r < L || l > R)
      u = r + 1;
    else {
      if (r >= fa[l] - n - 10)
        u = fa[l] - n - 10;
      else
        u = r + 1;
    }
    ans = 1ll * (u - l) * (u - l + 1) / 2 + s[r] - s[u - 1];
    printf("%I64d\n", ans);
  }
  return 0;
}
