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 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);
int n, m, q, to[N], f[N][20], mx[N][20], mn[N][20], depth[N];
bool vis[N];
long long ans[N];
vector<int> G[N];
vector<pair<int, int> > qus[N];
long long a[N << 2], lazy[N << 2];
inline void pull(int rt) { a[rt] = a[rt << 1] + a[rt << 1 | 1]; }
inline void push(int rt, int l, int mid, int r) {
  if (lazy[rt]) {
    a[rt << 1] += lazy[rt] * (mid - l + 1);
    a[rt << 1 | 1] += lazy[rt] * (r - mid);
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    lazy[rt] = 0;
  }
}
void update(int L, int R, long long val, int l, int r, int rt) {
  if (l >= L && r <= R) {
    a[rt] += val * (r - l + 1);
    lazy[rt] += val;
    return;
  }
  int mid = l + r >> 1;
  push(rt, l, mid, r);
  if (L <= mid) update(L, R, val, l, mid, rt << 1);
  if (R > mid) update(L, R, val, mid + 1, r, rt << 1 | 1);
  pull(rt);
}
long long query(int L, int R, int l, int r, int rt) {
  if (l >= L && r <= R) return a[rt];
  int mid = l + r >> 1;
  push(rt, l, mid, r);
  if (R <= mid)
    return query(L, R, l, mid, rt << 1);
  else if (L > mid)
    return query(L, R, mid + 1, r, rt << 1 | 1);
  else
    return query(L, R, l, mid, rt << 1) + query(L, R, mid + 1, r, rt << 1 | 1);
}
void dfs(int u, int fa) {
  vis[u] = true;
  depth[u] = depth[fa] + 1;
  f[u][0] = fa;
  mx[u][0] = u, mn[u][0] = u;
  for (int i = 1; i < 20; i++) {
    f[u][i] = f[f[u][i - 1]][i - 1];
    mx[u][i] = max(mx[u][i - 1], mx[f[u][i - 1]][i - 1]);
    mn[u][i] = min(mn[u][i - 1], mn[f[u][i - 1]][i - 1]);
  }
  for (auto& v : G[u]) {
    if (v == fa || vis[v]) continue;
    dfs(v, u);
  }
}
void getVal(int u, int v) {
  int MX = v, MN = v;
  for (int i = 19; i >= 0; i--) {
    if (depth[u] - depth[v] >= (1 << i)) {
      MX = max(MX, mx[u][i]);
      MN = min(MN, mn[u][i]);
      u = f[u][i];
    }
  }
  to[MN] = MX - 1;
}
void getTo(int u, int fa) {
  vis[u] = true;
  for (auto& v : G[u]) {
    if (v == fa)
      continue;
    else if (!vis[v])
      getTo(v, u);
    else if (vis[v]) {
      if (depth[u] > depth[v]) getVal(u, v);
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) to[i] = n;
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 0);
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++)
    if (!vis[i]) getTo(i, 0);
  for (int i = n - 1; i >= 1; i--) to[i] = min(to[i], to[i + 1]);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int L, R;
    scanf("%d%d", &L, &R);
    qus[L].push_back(make_pair(R, i));
  }
  for (int i = n; i >= 1; i--) {
    update(i, to[i], 1, 1, n, 1);
    for (auto& Q : qus[i]) ans[Q.second] = query(i, Q.first, 1, n, 1);
  }
  for (int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
  return 0;
}
