#include <bits/stdc++.h>
using namespace std;
template <class Q>
Q _sqr(Q x) {
  return x * x;
}
template <class Q>
void gi(Q &x) {
  char ch = getchar();
  x = 0;
  int s = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') s = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  if (s == -1) x = -x;
}
template <class Q>
void pi(Q x) {
  if (x < 0) {
    putchar('-');
    pi(-x);
  } else {
    if (x > 9) pi(x / 10);
    putchar(x % 10 + 48);
  }
}
const int sz = 100200;
const int huge = 1000000000;
const long long int hugell = 1ll << 61;
const double hugelf = 1e50;
const double eps = 1e-10;
const int ljz = 1000000007;
void MOD(int &x) {
  if (x >= ljz) x -= ljz;
  if (x < 0) x += ljz;
}
int Tm(int a, int b) {
  long long int x = a, y = b;
  return (x * y) % ljz;
}
int fpm(int x, int y) {
  int p = 1, s = x;
  for (; y; y >>= 1) {
    if (y & 1) p = Tm(p, s);
    s = Tm(s, s);
  }
  return p;
}
int Tm(int a, int b, int c) { return Tm(Tm(a, b), c); }
int Tm(int a, int b, int c, int d) { return Tm(Tm(a, b, c), d); }
int Tm(int a, int b, int c, int d, int e) { return Tm(Tm(a, b, c, d), e); }
int Tm(int a, int b, int c, int d, int e, int f) {
  return Tm(Tm(a, b, c, d, e), f);
}
int n, k;
long long int l;
int a[1002000];
pair<int, int> sa[1002000];
int f[1002000], s[1002000];
int N, O;
int main() {
  int i, j, p, q, r, ans = 0;
  gi(n);
  gi(l);
  gi(k);
  for (i = 0; i < n; i++) {
    gi(a[i]);
    f[i] = 1;
    sa[i] = make_pair(a[i], i);
  }
  sort(sa, sa + n);
  N = n;
  O = 0;
  for (i = 1; i < k; i++) {
    p = 0;
    r = 0;
    for (j = 0; j < n; j++) {
      while (p < n && sa[p].first <= sa[j].first)
        MOD(r += f[O + sa[p].second]), p++;
      f[N + sa[j].second] = r;
    }
    O = N;
    N += n;
  }
  for (i = 1; i <= k; i++) {
    if ((i - 1) * n + 1 > l) break;
    for (j = 0; j < n; j++) {
      long long int P = ((l / n) - i + 1);
      if (j < (l % n)) P++;
      MOD(ans += Tm(f[(i - 1) * n + j], P % ljz));
    }
  }
  printf("%d", ans);
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

<image>

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that: 

  * 1 ≤ x ≤ k
  * For each 1 ≤ j ≤ x - 1, <image>
  * For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing. 



Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1). 

Output

Print the answer modulo 1 000 000 007 in one line.

Examples

Input

3 5 3
5 9 1


Output

10


Input

5 10 3
1 2 3 4 5


Output

25

Note

In the first sample case, <image>. So all such sequences are: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class Q>
Q _sqr(Q x) {
  return x * x;
}
template <class Q>
void gi(Q &x) {
  char ch = getchar();
  x = 0;
  int s = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') s = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  if (s == -1) x = -x;
}
template <class Q>
void pi(Q x) {
  if (x < 0) {
    putchar('-');
    pi(-x);
  } else {
    if (x > 9) pi(x / 10);
    putchar(x % 10 + 48);
  }
}
const int sz = 100200;
const int huge = 1000000000;
const long long int hugell = 1ll << 61;
const double hugelf = 1e50;
const double eps = 1e-10;
const int ljz = 1000000007;
void MOD(int &x) {
  if (x >= ljz) x -= ljz;
  if (x < 0) x += ljz;
}
int Tm(int a, int b) {
  long long int x = a, y = b;
  return (x * y) % ljz;
}
int fpm(int x, int y) {
  int p = 1, s = x;
  for (; y; y >>= 1) {
    if (y & 1) p = Tm(p, s);
    s = Tm(s, s);
  }
  return p;
}
int Tm(int a, int b, int c) { return Tm(Tm(a, b), c); }
int Tm(int a, int b, int c, int d) { return Tm(Tm(a, b, c), d); }
int Tm(int a, int b, int c, int d, int e) { return Tm(Tm(a, b, c, d), e); }
int Tm(int a, int b, int c, int d, int e, int f) {
  return Tm(Tm(a, b, c, d, e), f);
}
int n, k;
long long int l;
int a[1002000];
pair<int, int> sa[1002000];
int f[1002000], s[1002000];
int N, O;
int main() {
  int i, j, p, q, r, ans = 0;
  gi(n);
  gi(l);
  gi(k);
  for (i = 0; i < n; i++) {
    gi(a[i]);
    f[i] = 1;
    sa[i] = make_pair(a[i], i);
  }
  sort(sa, sa + n);
  N = n;
  O = 0;
  for (i = 1; i < k; i++) {
    p = 0;
    r = 0;
    for (j = 0; j < n; j++) {
      while (p < n && sa[p].first <= sa[j].first)
        MOD(r += f[O + sa[p].second]), p++;
      f[N + sa[j].second] = r;
    }
    O = N;
    N += n;
  }
  for (i = 1; i <= k; i++) {
    if ((i - 1) * n + 1 > l) break;
    for (j = 0; j < n; j++) {
      long long int P = ((l / n) - i + 1);
      if (j < (l % n)) P++;
      MOD(ans += Tm(f[(i - 1) * n + j], P % ljz));
    }
  }
  printf("%d", ans);
  return 0;
}

```