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The **appeal** of a string is the number of **distinct** characters found in the string.

- For example, the appeal of
`"abbca"`

is`3`

because it has`3`

distinct characters:`'a'`

,`'b'`

, and`'c'`

.

Given a string `s`

, return *the total appeal of all of its substrings.*

A **substring** is a contiguous sequence of characters within a string.

**Example 1:**

```
Input: s = "abbca"
Output: 28
Explanation: The following are the substrings of "abbca":
- Substrings of length 1: "a", "b", "b", "c", "a" have an appeal of 1, 1, 1, 1, and 1 respectively. The sum is 5.
- Substrings of length 2: "ab", "bb", "bc", "ca" have an appeal of 2, 1, 2, and 2 respectively. The sum is 7.
- Substrings of length 3: "abb", "bbc", "bca" have an appeal of 2, 2, and 3 respectively. The sum is 7.
- Substrings of length 4: "abbc", "bbca" have an appeal of 3 and 3 respectively. The sum is 6.
- Substrings of length 5: "abbca" has an appeal of 3. The sum is 3.
The total sum is 5 + 7 + 7 + 6 + 3 = 28.
```

**Example 2:**

```
Input: s = "code"
Output: 20
Explanation: The following are the substrings of "code":
- Substrings of length 1: "c", "o", "d", "e" have an appeal of 1, 1, 1, and 1 respectively. The sum is 4.
- Substrings of length 2: "co", "od", "de" have an appeal of 2, 2, and 2 respectively. The sum is 6.
- Substrings of length 3: "cod", "ode" have an appeal of 3 and 3 respectively. The sum is 6.
- Substrings of length 4: "code" has an appeal of 4. The sum is 4.
The total sum is 4 + 6 + 6 + 4 = 20.
```

**Constraints:**

`1 <= s.length <= 10`

^{5}`s`

consists of lowercase English letters.

```
public long appealSum(String s) {
int last[] = new int[26];
long res = 0;
for (int i = 0; i < s.length(); ++i) {
last[s.charAt(i) - 'a'] = i + 1;
for (int j: last)
res += j;
}
return res;
}
```

```
long long appealSum(string s) {
vector<int> last(26);
long res = 0, n = s.size();
for (int i = 0; i < n; ++i) {
last[s[i] - 'a'] = i + 1;
for (int j: last)
res += j;
}
return res;
}
```

```
def appealSum(self, s):
last = {}
res = 0
for i,c in enumerate(s):
last[c] = i + 1
res += sum(last.values())
return res
```

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