How to declare a function interface with multiple call options?

Question

typescript

How to declare a function interface with multiple call options?
The function must take either one or 4 parameters
I try like this

export interface IValidator {
    (value: any): ValidationError;
    (value: any, instance: any, name: string, strict: boolean): ValidationError;
}

However, the following code throws an error when declaring BadValidator

const BadValidator: IValidator = (value: any, i: any, n: string, s: boolean): ValidationError => {
    return null;
};
const SmallValidator: IValidator = (value: any): ValidationError => {
    return null;
};

Type '(value: any, i: any, n: string, s: boolean) => ValidationError' 
is not assignable to type 'IValidator'.