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 = 300011;
int now[maxn], pre[maxn << 1], son[maxn << 1], tot = 0;
void add(int a, int b) {
  pre[++tot] = now[a];
  now[a] = tot;
  son[tot] = b;
}
void cc(int a, int b) {
  add(a, b);
  add(b, a);
}
int n, m;
void init() {
  scanf("%d%d", &n, &m);
  for (int i = 1, a, b; i <= m; ++i) scanf("%d%d", &a, &b), cc(a, b);
}
struct Tp {
  int l, r;
  Tp(int l_ = 0, int r_ = 0) : l(l_), r(r_) {}
} p[maxn];
int dfn[maxn], low[maxn], sign = 0, Stack[maxn], stot = 0, N = 0;
void tarjan(int x, int fa) {
  dfn[x] = low[x] = ++sign;
  Stack[++stot] = x;
  for (int p = now[x]; p; p = pre[p])
    if (son[p] != fa) {
      if (!dfn[son[p]])
        tarjan(son[p], x), low[x] = min(low[x], low[son[p]]);
      else
        low[x] = min(low[x], dfn[son[p]]);
    }
  if (dfn[x] == low[x]) {
    int mn = n + 1, mx = 0;
    Stack[stot + 1] = 0;
    while (Stack[stot + 1] != x) {
      mn = min(mn, Stack[stot]);
      mx = max(mx, Stack[stot--]);
    }
    ::p[++N] = Tp(mn, mx);
  }
}
inline bool cmp_Tp_l(Tp a, Tp b) { return a.l < b.l; }
void work() {
  memset(dfn, 0, sizeof(dfn));
  for (int i = 1; i <= n; ++i)
    if (!dfn[i]) tarjan(i, 0);
  sort(p + 1, p + N + 1, cmp_Tp_l);
  static int a[maxn];
  static long long s[maxn];
  for (int l = n, i = N, mn = n + 1; l >= 1; --l) {
    while (i >= 1 && p[i].l == l) {
      if (p[i].l == p[i].r) {
        --i;
        continue;
      }
      mn = min(mn, p[i--].r);
    }
    a[l] = mn;
  }
  s[0] = 0;
  for (int i = 1; i <= n; ++i) s[i] = s[i - 1] + a[i];
  int Q;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    int R = upper_bound(a + 1, a + n + 1, r) - a - 1;
    long long ans = 0;
    if (R >= l) ans = (r + 1LL) * (R - l + 1) - (s[R] - s[l - 1]);
    ans = (r - l + 1LL) * (r - l + 2) / 2 - ans;
    printf("%lld\n", ans);
  }
}
int main() {
  init();
  work();
  return 0;
}
