#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 1e6 + 10;
long long l;
int cnt = 0;
int n, k;
int a[N], b[N];
vector<long long> f[N], g[N];
long long c[N];
long long m;
long long ret = 0;
void update(int x, long long val) {
  for (; x <= n; x += x & -x) (c[x] += val) %= mod;
}
long long query(int x) {
  long long ret = 0;
  for (; x; x -= x & -x) (ret += c[x]) %= mod;
  return ret;
}
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i(1); i <= (n); ++i) scanf("%d", a + i), b[i] = a[i];
  sort(b + 1, b + n + 1);
  cnt = unique(b + 1, b + n + 1) - b - 1;
  for (int i(1); i <= (n); ++i)
    a[i] = lower_bound(b + 1, b + cnt + 1, a[i]) - b;
  for (int i(0); i <= (n + 1); ++i) f[i].resize(k + 3), g[i].resize(k + 3);
  for (int i(1); i <= (n); ++i) f[i][1] = 1, update(a[i], 1);
  for (int j(2); j <= (k); ++j) {
    for (int i(1); i <= (n); ++i) f[i][j] = query(a[i]);
    for (int i(0); i <= (n + 1); ++i) c[i] = 0;
    for (int i(1); i <= (n); ++i) update(a[i], f[i][j]);
  }
  for (int i(1); i <= (n); ++i) g[i][1] = f[i][1];
  for (int j(2); j <= (k); ++j)
    for (int i(1); i <= (n); ++i) g[i][j] = (g[i][j - 1] + f[i][j]) % mod;
  m = (l - 1) / (long long)n;
  l -= m * n;
  if (m < k) {
    long long ans = 0;
    for (int i(1); i <= (n); ++i)
      for (int j(1); j <= (m); ++j) ans = (ans + g[i][j]) % mod;
    for (int i(1); i <= (l); ++i) ans = (ans + g[i][m + 1]) % mod;
    printf("%lld\n", ans);
    return 0;
  }
  for (int i(1); i <= (n); ++i)
    for (int j(1); j <= (k); ++j) ret = (ret + g[i][j]) % mod;
  long long mul = 0;
  for (int i(1); i <= (n); ++i) mul = (mul + g[i][k]) % mod;
  m = (m - k + mod) % mod;
  mul = (mul * m) % mod;
  ret = (ret + mul) % mod;
  for (int i(1); i <= (l); ++i) ret = (ret + g[i][k]) % mod;
  printf("%lld\n", ret);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
const long long mod = 1e9 + 7;
const int N = 1e6 + 10;
long long l;
int cnt = 0;
int n, k;
int a[N], b[N];
vector<long long> f[N], g[N];
long long c[N];
long long m;
long long ret = 0;
void update(int x, long long val) {
  for (; x <= n; x += x & -x) (c[x] += val) %= mod;
}
long long query(int x) {
  long long ret = 0;
  for (; x; x -= x & -x) (ret += c[x]) %= mod;
  return ret;
}
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i(1); i <= (n); ++i) scanf("%d", a + i), b[i] = a[i];
  sort(b + 1, b + n + 1);
  cnt = unique(b + 1, b + n + 1) - b - 1;
  for (int i(1); i <= (n); ++i)
    a[i] = lower_bound(b + 1, b + cnt + 1, a[i]) - b;
  for (int i(0); i <= (n + 1); ++i) f[i].resize(k + 3), g[i].resize(k + 3);
  for (int i(1); i <= (n); ++i) f[i][1] = 1, update(a[i], 1);
  for (int j(2); j <= (k); ++j) {
    for (int i(1); i <= (n); ++i) f[i][j] = query(a[i]);
    for (int i(0); i <= (n + 1); ++i) c[i] = 0;
    for (int i(1); i <= (n); ++i) update(a[i], f[i][j]);
  }
  for (int i(1); i <= (n); ++i) g[i][1] = f[i][1];
  for (int j(2); j <= (k); ++j)
    for (int i(1); i <= (n); ++i) g[i][j] = (g[i][j - 1] + f[i][j]) % mod;
  m = (l - 1) / (long long)n;
  l -= m * n;
  if (m < k) {
    long long ans = 0;
    for (int i(1); i <= (n); ++i)
      for (int j(1); j <= (m); ++j) ans = (ans + g[i][j]) % mod;
    for (int i(1); i <= (l); ++i) ans = (ans + g[i][m + 1]) % mod;
    printf("%lld\n", ans);
    return 0;
  }
  for (int i(1); i <= (n); ++i)
    for (int j(1); j <= (k); ++j) ret = (ret + g[i][j]) % mod;
  long long mul = 0;
  for (int i(1); i <= (n); ++i) mul = (mul + g[i][k]) % mod;
  m = (m - k + mod) % mod;
  mul = (mul * m) % mod;
  ret = (ret + mul) % mod;
  for (int i(1); i <= (l); ++i) ret = (ret + g[i][k]) % mod;
  printf("%lld\n", ret);
  return 0;
}

```