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>
const int MAXN = 1 << 12;
const int MAXM = 1 << 16;
const int B = 1 << 8;
int n, q, a[MAXN + 5], bel[MAXN + 5], ps[MAXN + 5];
std::vector<std::pair<int, int> > v;
inline int merge(int x, int y) {
  if (!x) return y;
  if (!y) return x;
  v.push_back(std::make_pair(x, y));
  return v.size() + n;
}
struct Node {
  std::vector<int> S;
  int ps[B][B];
  Node(int x = -1) {
    if (x != -1) S.push_back(::ps[x]), ps[0][0] = ::ps[x] + 1;
  }
  inline int query(int l, int r) {
    l = std::lower_bound(S.begin(), S.end(), l) - S.begin();
    r = std::upper_bound(S.begin(), S.end(), r) - S.begin() - 1;
    if (l > r) return 0;
    return ps[l][r];
  }
  friend Node operator+(Node A, Node B) {
    Node C;
    C.S.resize(A.S.size() + B.S.size());
    merge(A.S.begin(), A.S.end(), B.S.begin(), B.S.end(), C.S.begin());
    for (int i = 0; i <= (int)C.S.size() - 1; ++i) {
      for (int j = i; j <= (int)C.S.size() - 1; ++j) {
        int l = C.S[i], r = C.S[j];
        C.ps[i][j] = merge(A.query(l, r), B.query(l, r));
      }
    }
    return C;
  }
} b[(MAXN / B) + 5];
inline Node work(int l, int r) {
  if (l == r) return Node(l);
  int mid = (l + r) >> 1;
  return work(l, mid) + work(mid + 1, r);
}
int ans[MAXM + 5];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i <= n - 1; ++i) scanf("%d", a + i), a[i]--, ps[a[i]] = i;
  for (int i = 0; i <= n - 1; ++i) bel[i] = i / B;
  int m = bel[n - 1];
  for (int i = 0; i <= m; ++i)
    b[i] = work(i * B, std::min(n - 1, (i + 1) * B - 1));
  for (int i = 1; i <= q; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l;
    --r;
    int res = 0;
    for (int i = 0; i <= m; ++i) res = merge(res, b[i].query(l, r));
    ans[i] = res;
  }
  printf("%d\n", n + (int)v.size());
  for (auto x : v) printf("%d %d\n", x.first, x.second);
  for (int i = 1; i <= q; ++i) printf("%d ", ans[i]);
  puts("");
  return 0;
}