DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
constexpr long double pi{3.1415926535897932384};
constexpr long long mod{1000000007};
long long power(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
const long long N = 100005;
long long a[N], color[4 * N], tree[4 * N], lazy[4 * N];
void inline merge(long long node) {
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
void build(long long node, long long start, long long end) {
  if (start == end) {
    color[node] = start;
    return;
  }
  if (start > end) return;
  long long mid = (start + end) >> 1;
  build(2 * node, start, mid);
  build(2 * node + 1, mid + 1, end);
}
void lazyupdate(long long node, long long start, long long end) {
  if (lazy[node] != 0) {
    long long dx = lazy[node];
    lazy[node] = 0;
    tree[node] += dx * (end - start + 1);
    if (start != end) {
      lazy[2 * node] += dx;
      lazy[2 * node + 1] += dx;
    }
  }
}
void godown(long long node, long long start, long long end, long long val) {
  lazyupdate(node, start, end);
  if (color[node] > 0) {
    lazy[node] += abs(color[node] - val);
    lazyupdate(node, start, end);
    color[node] = 0;
  } else {
    long long mid = (start + end) >> 1;
    godown(2 * node, start, mid, val);
    godown(2 * node + 1, mid + 1, end, val);
    merge(node);
  }
}
void update(long long node, long long start, long long end, long long l,
            long long r, long long val) {
  lazyupdate(node, start, end);
  if (start > r || end < l || l > r) return;
  if (start >= l && end <= r) {
    godown(node, start, end, val);
    color[node] = val;
    return;
  }
  long long mid = (start + end) >> 1;
  if (color[node] > 0) {
    color[2 * node] = color[node];
    color[2 * node + 1] = color[node];
    color[node] = 0;
  }
  update(2 * node, start, mid, l, r, val);
  update(2 * node + 1, mid + 1, end, l, r, val);
  merge(node);
}
long long find(long long node, long long start, long long end, long long l,
               long long r) {
  lazyupdate(node, start, end);
  if (l <= start && end <= r) {
    return tree[node];
  }
  if (start > end || l > end || r < start) {
    return 0;
  }
  long long mid = (start + end) >> 1;
  return find(2 * node, start, mid, l, r) +
         find(2 * node + 1, mid + 1, end, l, r);
}
void solve() {
  long long n, m;
  cin >> n >> m;
  build(1, 1, n);
  while (m--) {
    long long a, b, c;
    cin >> a >> b >> c;
    if (a == 1) {
      long long val;
      cin >> val;
      update(1, 1, n, b, c, val);
    } else {
      cout << find(1, 1, n, b, c) << '\n';
    }
  }
}
signed main() {
  long long t{1};
  while (t--) {
    solve();
  }
  return 0;
}
