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 maxn = 500010, inf = 1e9;
struct poi {
  int too, pre;
} e[maxn << 1];
struct tjm {
  int mx, mn;
} q[maxn];
int n, m, x, y, tot, tott, top, color, L, R, Q;
int last[maxn], dfn[maxn], low[maxn], st[maxn], lack[maxn], mx[maxn], mn[maxn],
    col[maxn], nxt[maxn];
long long sum[maxn], nxtsum[maxn];
inline void read(int &k) {
  int f = 1;
  k = 0;
  char c = getchar();
  while (c < '0' || c > '9') c == '-' && (f = -1), c = getchar();
  while (c <= '9' && c >= '0') k = k * 10 + c - '0', c = getchar();
  k *= f;
}
inline void add(int x, int y) {
  e[++tot] = (poi){y, last[x]};
  last[x] = tot;
}
void tarjan(int x, int fa) {
  dfn[x] = low[x] = ++tott;
  st[++top] = x;
  lack[x] = top;
  for (int i = last[x], too; i; i = e[i].pre)
    if ((too = e[i].too) != fa) {
      if (!dfn[too = e[i].too])
        tarjan(too, x), low[x] = min(low[x], low[too]);
      else if (!col[too])
        low[x] = min(low[x], dfn[too]);
    }
  if (dfn[x] == low[x])
    for (q[++color].mn = inf; lack[x] <= top; top--) {
      col[st[top]] = color;
      q[color].mx = max(q[color].mx, st[top]);
      q[color].mn = min(q[color].mn, st[top]);
    }
}
inline bool cmp(tjm a, tjm b) { return a.mx < b.mx; }
int main() {
  read(n);
  read(m);
  for (int i = 1; i <= m; i++) read(x), read(y), add(x, y), add(y, x);
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) tarjan(i, 0);
  sort(q + 1, q + 1 + color, cmp);
  int j = 1;
  for (int i = 1; i <= color; i++)
    if (q[i].mn != q[i].mx)
      for (; j <= q[i].mn; j++) nxt[j] = q[i].mx - 1;
  for (int i = j; i <= n; i++) nxt[i] = n;
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + nxt[i] - i + 1;
  read(Q);
  for (int i = 1; i <= Q; i++) {
    read(L);
    read(R);
    int l = L - 1, r = R;
    while (l < r) {
      int mid = (l + r + 1) >> 1;
      if (nxt[mid] <= R)
        l = mid;
      else
        r = mid - 1;
    }
    printf("%lld\n", sum[l] - sum[L - 1] + 1ll * (R + 1) * (R - l) -
                         (1ll * (l + 1 + R) * (R - l) >> 1));
  }
}
