#include <bits/stdc++.h>
using namespace std;
long long n, l, k;
long long a[1000006];
long long f[1000005];
int nn;
long long c[1000005];
long long ans;
long long dq;
long long s[1000005];
const long long mod = 1e9 + 7;
int lowbit(int x) { return x & (-x); }
long long sum(int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= lowbit(i)) ans += c[i], ans %= mod;
  return ans;
}
void add(int x, long long w) {
  for (int i = x; i <= 1000000; i += lowbit(i)) c[i] += w, c[i] %= mod;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
  for (int i = 1; i <= n; ++i) f[i] = a[i];
  sort(f + 1, f + n + 1);
  nn = unique(f + 1, f + n + 1) - (f + 1);
  for (int i = 1; i <= n; ++i) a[i] = lower_bound(f + 1, f + nn + 1, a[i]) - f;
  memset(f, 0, sizeof(f));
  memset(c, 0, sizeof(c));
  long long maxx = n * k;
  for (int i = 1; i <= maxx; ++i) {
    if (i <= n) {
      f[i] = 1;
      s[1]++;
      add(a[i], 1);
      continue;
    }
    dq = a[(i - 1) % n + 1];
    f[i] = sum(dq), f[i] %= mod;
    s[(i - 1) / n + 1] = (s[(i - 1) / n + 1] + f[i]) % mod;
    if (i % n == 0) {
      for (int j = 0; j <= n; ++j) c[j] = 0;
      for (int j = i - n + 1; j <= i; ++j) add(a[j - (i - n)], f[j]);
    }
  }
  long long m = l / n;
  ans = 0;
  for (int i = 1; i <= k; ++i) {
    if (m < i) break;
    dq = (m - i + 1) % mod;
    dq = dq * s[i] % mod;
    ans = (ans + dq) % mod;
  }
  memset(f, 0, sizeof(f));
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= l % n; ++i) {
    ans++;
    ans %= mod;
    add(a[i], 1);
  }
  maxx = min(n * k - n, l - l % n);
  for (int i = 1; i <= maxx; ++i) {
    dq = a[(i - 1) % n + 1];
    f[i] = sum(nn) - sum(dq - 1), f[i] += mod, f[i] %= mod;
    ans = (ans + f[i]) % mod;
    if (i % n == 0) {
      for (int j = 0; j <= n; ++j) c[j] = 0;
      for (int j = i - n + 1; j <= i; ++j) add(a[j - (i - n)], f[j]);
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
long long n, l, k;
long long a[1000006];
long long f[1000005];
int nn;
long long c[1000005];
long long ans;
long long dq;
long long s[1000005];
const long long mod = 1e9 + 7;
int lowbit(int x) { return x & (-x); }
long long sum(int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= lowbit(i)) ans += c[i], ans %= mod;
  return ans;
}
void add(int x, long long w) {
  for (int i = x; i <= 1000000; i += lowbit(i)) c[i] += w, c[i] %= mod;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
  for (int i = 1; i <= n; ++i) f[i] = a[i];
  sort(f + 1, f + n + 1);
  nn = unique(f + 1, f + n + 1) - (f + 1);
  for (int i = 1; i <= n; ++i) a[i] = lower_bound(f + 1, f + nn + 1, a[i]) - f;
  memset(f, 0, sizeof(f));
  memset(c, 0, sizeof(c));
  long long maxx = n * k;
  for (int i = 1; i <= maxx; ++i) {
    if (i <= n) {
      f[i] = 1;
      s[1]++;
      add(a[i], 1);
      continue;
    }
    dq = a[(i - 1) % n + 1];
    f[i] = sum(dq), f[i] %= mod;
    s[(i - 1) / n + 1] = (s[(i - 1) / n + 1] + f[i]) % mod;
    if (i % n == 0) {
      for (int j = 0; j <= n; ++j) c[j] = 0;
      for (int j = i - n + 1; j <= i; ++j) add(a[j - (i - n)], f[j]);
    }
  }
  long long m = l / n;
  ans = 0;
  for (int i = 1; i <= k; ++i) {
    if (m < i) break;
    dq = (m - i + 1) % mod;
    dq = dq * s[i] % mod;
    ans = (ans + dq) % mod;
  }
  memset(f, 0, sizeof(f));
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= l % n; ++i) {
    ans++;
    ans %= mod;
    add(a[i], 1);
  }
  maxx = min(n * k - n, l - l % n);
  for (int i = 1; i <= maxx; ++i) {
    dq = a[(i - 1) % n + 1];
    f[i] = sum(nn) - sum(dq - 1), f[i] += mod, f[i] %= mod;
    ans = (ans + f[i]) % mod;
    if (i % n == 0) {
      for (int j = 0; j <= n; ++j) c[j] = 0;
      for (int j = i - n + 1; j <= i; ++j) add(a[j - (i - n)], f[j]);
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```