Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0;
  bool f = 1;
  char s = getchar();
  while ((s < '0' || s > '9') && s > 0) {
    if (s == '-') f ^= 1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') x = (x << 1) + (x << 3) + s - '0', s = getchar();
  return f ? x : -x;
}
int n, q, a[4101], IN, ans[65541], B[20][4101];
struct itn {
  int l, r;
  itn() {}
  itn(int L, int R) { l = L, r = R; }
  bool operator<(const itn& b) const {
    if (r - l != b.r - b.l)
      return r - l < b.r - b.l;
    else
      return l < b.l;
  }
} e[65541];
vector<itn> cz;
struct tre {
  map<itn, int> mp;
} t[4101 * 3];
inline void build(int x, int l, int r, int dep) {
  if (l == r) {
    int u = B[dep][l] = B[0][l];
    t[x].mp[itn(u, u)] = u;
    return;
  }
  int mid = (l + r) >> 1, u = l, v = mid + 1, w = l;
  build(x << 1, l, mid, dep + 1), build(x << 1 | 1, mid + 1, r, dep + 1);
  while (u <= mid && v <= r) {
    if (B[dep + 1][u] < B[dep + 1][v])
      B[dep][w] = B[dep + 1][u], u++, w++;
    else
      B[dep][w] = B[dep + 1][v], v++, w++;
  }
  while (u <= mid) B[dep][w] = B[dep + 1][u], u++, w++;
  while (v <= r) B[dep][w] = B[dep + 1][v], v++, w++;
}
inline int query(int x, int l, int r, int a, int b, int dep) {
  int mid = (l + r) >> 1, c = B[dep][l], d = B[dep][r];
  if (c > b || d < a) return 0;
  a = max(a, c), b = min(b, d);
  if (a > c || b < d) {
    int u = l - 1, v = l - 1;
    for (int i = 13 - dep; i >= 0; i--)
      if (u + (1 << i) < r && B[dep][u + (1 << i)] < a) u += (1 << i);
    u++;
    for (int i = 13 - dep; i >= 0; i--)
      if (v + (1 << i) <= r && B[dep][v + (1 << i)] <= b) v += (1 << i);
    a = B[dep][u], b = B[dep][v];
  }
  if (dep > 0) {
    int o = t[x].mp[itn(a, b)];
    if (o > 0) return o;
  }
  if (l == r) return t[x].mp[itn(c, d)];
  int u = query(x << 1, l, mid, a, b, dep + 1),
      v = query(x << 1 | 1, mid + 1, r, a, b, dep + 1);
  if (!u || !v) return u | v;
  cz.push_back(itn(u, v)), IN++;
  if (dep > 0) t[x].mp[itn(a, b)] = IN;
  return IN;
}
int main() {
  n = read(), q = read(), IN = n;
  for (int i = 1; i <= n; i++) a[i] = read(), B[0][a[i]] = i;
  for (int i = 1; i <= q; i++) e[i].l = read(), e[i].r = read();
  build(1, 1, n, 1);
  for (int i = 1; i <= q; i++) ans[i] = query(1, 1, n, e[i].l, e[i].r, 1);
  printf("%d\n", IN);
  for (int i = 0; i < cz.size(); i++) printf("%d %d\n", cz[i].l, cz[i].r);
  for (int i = 1; i <= q; i++) printf("%d ", ans[i]);
  putchar('\n');
  return 0;
}