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 = 3e5 + 1;
int ch[MAXN + 10][2], Fa[MAXN + 10], rev[MAXN + 10], que[MAXN + 10];
bool isroot(int u) {
  if (!Fa[u]) return 1;
  return ch[Fa[u]][0] != u && ch[Fa[u]][1] != u;
}
void Rotate(int u) {
  bool d = ch[Fa[u]][1] == u;
  int x = Fa[u], y = Fa[x];
  Fa[u] = y;
  if (!isroot(x)) ch[y][ch[y][1] == x] = u;
  ch[x][d] = ch[u][!d];
  Fa[ch[u][!d]] = x;
  ch[u][!d] = x;
  Fa[x] = u;
}
void push_down(int u) {
  if (rev[u]) {
    rev[ch[u][0]] ^= 1;
    swap(ch[ch[u][0]][0], ch[ch[u][0]][1]);
    rev[ch[u][1]] ^= 1;
    swap(ch[ch[u][1]][0], ch[ch[u][1]][1]);
    rev[u] ^= 1;
  }
}
void Splay(int u) {
  int top = 0;
  que[++top] = u;
  for (int i = u; !isroot(i); i = Fa[i]) que[++top] = Fa[i];
  top++;
  while (--top) push_down(que[top]);
  while (!isroot(u)) {
    int x = Fa[u], y = Fa[x];
    if (!isroot(x)) {
      if ((ch[y][0] == x) ^ (ch[x][0] == u))
        Rotate(x);
      else
        Rotate(u);
    }
    Rotate(u);
  }
}
void Access(int u) {
  int t = 0;
  while (u) {
    Splay(u);
    ch[u][1] = t;
    t = u;
    u = Fa[u];
  }
}
void reset_root(int u) {
  Access(u);
  Splay(u);
  rev[u] ^= 1;
  swap(ch[u][0], ch[u][1]);
}
void Link(int x, int y) {
  reset_root(x);
  Fa[x] = y;
  Splay(x);
}
void Cut(int x, int y) {
  reset_root(x);
  Access(y);
  Splay(y);
  ch[y][0] = Fa[x] = 0;
}
int find_root(int x) {
  while (Fa[x]) x = Fa[x];
  return x;
}
int n, m, q;
vector<int> g[MAXN];
int reach[MAXN];
int up[MAXN];
long long sum[MAXN];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int nxt = 1;
  for (int i = 1; i <= n; ++i) {
    if (nxt < i) nxt = i;
    while (nxt <= n) {
      int j;
      bool work = true;
      for (j = 0; j < g[nxt].size(); ++j) {
        int e = g[nxt][j];
        if (e < i || e >= nxt) continue;
        if (find_root(e) == find_root(nxt)) {
          work = false;
          while (j) {
            --j;
            int e = g[nxt][j];
            if (e < i || e >= nxt) continue;
            Cut(e, nxt);
          }
          break;
        } else {
          Link(e, nxt);
        }
      }
      if (work)
        ++nxt;
      else
        break;
    }
    reach[i] = nxt - 1;
    for (auto e : g[i]) {
      if (e < i || e >= nxt) continue;
      Cut(e, i);
    }
  }
  int last = 1;
  for (int i = 1; i <= n; ++i) {
    while (last <= n && reach[last] <= i) ++last;
    up[i] = last;
  }
  sum[0] = 0;
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + reach[i] - i + 1;
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    int l, r;
    scanf("%d %d", &l, &r);
    int tmp = up[r] - 1;
    tmp = min(r, tmp);
    tmp = max(tmp, l - 1);
    long long left = r - tmp;
    printf("%lld\n", sum[tmp] - sum[l - 1] + (1ll + left) * left / 2);
  }
}
