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>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
const long long MOD2 = (long long)1000000007 * (long long)1000000007;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 1000000007) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
const int B = 1 << 8;
int a[(1 << 12) + 5], pos[(1 << 12) + 5], sz;
vector<pair<int, int> > A;
int mg(int u, int v) {
  if (u == 0 || v == 0) return v | u;
  A.push_back({u, v});
  return ++sz;
}
int f[(B * 2) + 2][B + 2][B + 2];
struct mb {
  int L, R;
  int id[(1 << 12) + 5], val[B + 2][B + 2];
  void cal(int rt, int l, int r, vector<int>& vp) {
    if (l == r) {
      assert(vp.size() > 1);
      f[rt][1][1] = pos[vp[1]];
      return;
    }
    int mid = (l + r) / 2;
    vector<int> vl(1), vr(1);
    vector<int> ps(((int)vp.size()) + 2, 0), psr(((int)vp.size()) + 2, 0);
    for (int i = 1; i < vp.size(); i++) {
      if (vp[i] <= mid) {
        vl.push_back(vp[i]);
        ps[i]++;
      } else {
        vr.push_back(vp[i]);
        psr[i]++;
      }
    }
    for (int(i) = 1; (i) <= (int)(vp.size()); (i)++) {
      ps[i] += ps[i - 1];
      psr[i] += psr[i - 1];
    }
    cal((rt << 1), l, mid, vl);
    cal((rt << 1 | 1), mid + 1, r, vr);
    for (int i = 1; i < vp.size(); i++) {
      for (int j = i; j < vp.size(); j++) {
        int A = ps[i - 1] + 1;
        int B = ps[j];
        int C = psr[i - 1] + 1, D = psr[j];
        f[rt][i][j] = mg(f[(rt << 1)][A][B], f[(rt << 1 | 1)][C][D]);
      }
    }
  }
  void build() {
    vector<int> vp(1);
    for (int(i) = 1; (i) <= (int)(n); (i)++) {
      id[i] = id[i - 1];
      if (L <= a[i] && a[i] <= R) {
        id[i]++;
        vp.push_back(a[i]);
      }
    }
    cal(1, L, R, vp);
    for (int i = 1; i < vp.size(); i++) {
      for (int j = i; j < vp.size(); j++) {
        val[i][j] = f[1][i][j];
      }
    }
  }
  int qy(int l, int r) { return val[id[l - 1] + 1][id[r]]; }
} blk[17];
void fmain(int tid) {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    scanf("%d", a + i);
    pos[a[i]] = i;
  }
  sz = n;
  for (int i = 1; (i - 1) * B + 1 <= n; i++) {
    blk[i].L = (i - 1) * B + 1;
    blk[i].R = min(n, i * B);
    blk[i].build();
  }
  vector<int> ans(m + 2, 0);
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    int l, r;
    scanf("%d%d", &l, &r);
    for (int j = 1; (j - 1) * B + 1 <= n; j++) {
      ans[i] = mg(ans[i], blk[j].qy(l, r));
    }
  }
  printf("%d\n", sz);
  for (auto p : A) printf("%d %d\n", p.first, p.second);
  for (int(i) = 1; (i) <= (int)(m); (i)++) printf("%d ", ans[i]);
  puts("");
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}