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 = 3e5 + 5;
int n, m, t, q;
int dfn[N], low[N], head[N], bel[N], tot, cnt, num;
int maxx[N], minn[N], a[N], ans[N], vis[N];
long long sum[N];
bool in[N];
stack<int> S;
struct p {
  int next, to, id;
} edge[N * 2];
inline void add(int u, int v, int id) {
  edge[++tot] = {head[u], v, id};
  head[u] = tot;
}
void tarjan(int u, int id) {
  dfn[u] = low[u] = ++cnt;
  in[u] = 1, S.push(u);
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    if (id == edge[i].id) continue;
    if (!dfn[to]) {
      tarjan(to, edge[i].id);
      low[u] = min(low[u], low[to]);
    } else if (in[to])
      low[u] = min(low[u], dfn[to]);
  }
  if (low[u] == dfn[u]) {
    int tmp;
    num++;
    do {
      tmp = S.top();
      S.pop();
      bel[tmp] = num;
      vis[num]++;
      maxx[num] = max(maxx[num], tmp);
      minn[num] = min(minn[num], tmp);
      in[tmp] = 0;
    } while (tmp != u);
  }
}
void init() {
  cnt = tot = num = 0;
  memset(dfn, 0, sizeof(dfn));
  memset(minn, 0x3f3f3f3f, sizeof(minn));
  memset(low, 0, sizeof(low));
  memset(head, -1, sizeof(head));
  memset(in, 0, sizeof(in));
}
int main() {
  init();
  scanf("%d%d", &n, &m);
  int u, v;
  for (int i = (1); i <= (m); i++) {
    scanf("%d%d", &u, &v);
    add(u, v, i);
    add(v, u, i);
  }
  for (int i = (1); i <= (n); i++)
    if (!dfn[i]) tarjan(i, -1);
  for (int i = (1); i <= (n); i++) a[i] = n + 1;
  int nnn = n + 1;
  for (int i = (n); i >= (1); i--) {
    if (vis[bel[i]] == 1) {
      a[i] = nnn;
      continue;
    }
    if (minn[bel[i]] == i) nnn = min(nnn, maxx[bel[i]]);
    a[i] = nnn;
  }
  for (int i = (1); i <= (n); i++) {
    ans[i] = a[i] - i;
    sum[i] = sum[i - 1] + ans[i];
  }
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    int l = x, r = y, aaa = x - 1;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (a[mid] <= y) {
        l = mid + 1;
        aaa = mid;
      } else
        r = mid - 1;
    }
    long long Ans = sum[aaa] - sum[x - 1];
    Ans = Ans + 1ll * (y - aaa) * (y - aaa + 1) / 2;
    printf("%lld\n", Ans);
  }
  return 0;
}
