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 = 300010;
int n, m, e, tot;
int hed[N], ver[N << 1], nxt[N << 1], vis[N];
int dep[N], fa[N], mxr[N];
long long sum[N];
pair<int, int> cir[N];
void add(int x, int y) {
  ver[++e] = y;
  nxt[e] = hed[x];
  hed[x] = e;
}
void update(int x, int y) {
  if (x > y) return;
  int mi = min(x, y), mx = max(x, y);
  if (dep[x] < dep[y]) swap(x, y);
  while (dep[x] > dep[y]) x = fa[x], mi = min(mi, x), mx = max(mx, x);
  while (x != y) {
    x = fa[x];
    y = fa[y];
    mi = min(mi, x);
    mx = max(mx, x);
    mi = min(mi, y);
    mx = max(mx, y);
  }
  cir[++tot] = make_pair(mi, mx);
}
void dfs(int x) {
  dep[x] = dep[fa[x]] + 1;
  vis[x] = 1;
  for (int i = hed[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == fa[x]) continue;
    if (vis[y]) {
      update(x, y);
      continue;
    }
    fa[y] = x;
    dfs(y);
  }
}
int calc(int l, int r) {
  if (mxr[r] <= r) return r;
  if (mxr[l] > r) return l - 1;
  int ret = l - 1, R = r;
  while (l <= r) {
    int mid = l + r >> 1;
    if (mxr[mid] <= R)
      ret = max(ret, mid), l = mid + 1;
    else
      r = mid - 1;
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  sort(cir + 1, cir + 1 + tot);
  cout << endl;
  for (int i = n, j = tot + 1; i >= 1; i--) {
    if (i == n)
      mxr[i] = n;
    else
      mxr[i] = mxr[i + 1];
    while (j > 1 && cir[j - 1].first >= i)
      j--, mxr[i] = min(mxr[i], cir[j].second - 1);
  }
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + (mxr[i] - i + 1);
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int pos = calc(l, r);
    cout << (sum[pos] - sum[l - 1]) + (long long)(r - pos) * (r - pos + 1) / 2
         << endl;
  }
  return 0;
}
