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;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x > y ? y : x;
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= f;
}
struct Edge {
  int next, num;
} e[400010 * 3];
struct Info {
  int l, r;
  bool operator<(const Info &a) const { return l < a.l; }
} a[400010];
int l, cnt, tot, Time, f[400010], g[400010], in[400010], st[400010],
    dfn[400010], low[400010];
long long sum[400010];
void add(int x, int y) {
  e[++cnt] = (Edge){e[x].next, y};
  e[x].next = cnt;
}
void tarjan(int x, int fa) {
  dfn[x] = low[x] = ++Time;
  st[++l] = x, in[x] = 1;
  for (int p = e[x].next; p; p = e[p].next) {
    int k = e[p].num;
    if (k == fa) continue;
    if (!dfn[k])
      tarjan(k, x), chkmin(low[x], low[k]);
    else if (in[k])
      chkmin(low[x], dfn[k]);
  }
  if (dfn[x] == low[x]) {
    if (st[l] == x) return in[st[l--]] = 0, void();
    a[++tot] = (Info){x, x};
    while (st[l] != x)
      chkmin(a[tot].l, st[l]), chkmax(a[tot].r, st[l]), in[st[l--]] = 0;
    in[st[l--]] = 0;
  }
}
int main() {
  int n, m;
  read(n), read(m);
  cnt = n;
  for (int i = 1; i <= m; i++) {
    int x, y;
    read(x), read(y);
    add(x, y), add(y, x);
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) Time = l = 0, tarjan(i, 0);
  sort(a + 1, a + tot + 1);
  a[0].l = 0;
  g[tot] = a[tot].r;
  for (int i = tot - 1; i > 0; i--) g[i] = min(a[i].r, g[i + 1]);
  for (int i = 1; i <= n; i++) {
    if (i > a[tot].l) {
      f[i] = n;
      continue;
    }
    int l = 1, r = tot, ans = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (i <= a[mid].l)
        ans = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    f[i] = g[ans] - 1;
  }
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + f[i] - i + 1;
  int q;
  read(q);
  while (q--) {
    int l, r;
    read(l), read(r);
    int tl = l, tr = r, pos = tl - 1;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (f[mid] <= tr)
        pos = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    long long ans =
        sum[pos] - sum[tl - 1] + 1ll * (tr - pos + 1) * (tr - pos) / 2;
    cout << ans << "\n";
  }
  return 0;
}
